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given two parallel lines $P$, $Q$ and two points $X$,$Y$ how to construct a rhombus $ABCD$ passing through $X$,$Y$ and opposite sides lie on $P$ and $Q$.

I solved a special case of this problem where one point $Y$ lies on the parallel line. any hints or ideas?

If $Y$ lies on line $q$, draw a circle with center $Y$ and radius $XY$. The circle will intersect line $p$ at point $Z$. If I draw a perpendicular bisector of $ZX$ it will intersect line $p$ at $B$. Then it is easy to construct the required rhombus.

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  • $\begingroup$ Hello, it would help others save time if you could show your solution for the "special case" of your problem briefly; there might be similarities in the two processes. $\endgroup$ Commented Feb 8, 2021 at 5:26

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In a rhombus the distance between one pair of parallel sides is same as distance between another pair of parallel opposite sides.

So first find perpendicular distance $AB$ between $p$ and $q$.

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Next with $X$ as center and radius $AB$, draw a circle. From $Y$ drop tangents to this circle. Extend a suitable tangent to intersect $p,q$ in $Q$ and $R$.

Now construct a line through $X$ parallel to $QR$ to intersect $p,q$ in $P,S$. $PQRS$ is our desired rhombus.

We used the fact that $AB=XT$. For this construction, $X,Y$ need not lie between $p,q$.

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